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Let $f$ be continuous on $[a,b]$, where $a<b$, such that $f(x)≠0$ for all $x ∈ [a,b]$. Prove that there is a $c>0$, either $f(x)>c$ or $f(x)<-c$ for all $x ∈ [a,b]$.

What I did,

Proving by contradiction, Let $f(x)>c$ for some $x_{1} ∈ [a,b]$ and $f(x)<-c$ for some $x_{2} ∈ [a,b]$. By the Intermediate Value Theorem, there exists exists a number $z ∈ (a,b) $ such that $f(z) = 0$, however this contradicts the fact that $f(x)≠0$ for all $x ∈ [a,b]$. Hence $f(x)>c$ or $f(x)<-c$ for all $x ∈ [a,b]$.

Hi, was wondering whether my proof is sound. Do help me verify! Thanks!

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  • $\begingroup$ Looks good to me $\endgroup$ – Snufsan Sep 25 '16 at 11:19
  • $\begingroup$ Your contrary assumption is too strong. A function which is not constantly greater than some fixed positive value need not cross the axis. However z if there is no $c>0$ such that $f(x)>c$ for all $x$ then there is a sequence of points such that the function tends to zero along these points. $\endgroup$ – SamM Sep 25 '16 at 11:23
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You got the negation of the statment wrong. The negation of "There exists a $c > 0$ such that for all x $f(x) > c$ or for all x $f(x) < -c$" is actually "for each $c > 0$ there is an $x$ such that $f(x) < c$ and there is an $x$ such that $f(x) > -c$". That is the statement from which you should derive a contradiction.

Note: the point of the question is to use the fact that you are on a closed interval. On an open interval, for instance $(0,1)$, the statement does not hold (just look at $f(x) = 1-x$).

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  • $\begingroup$ Hi! Thank you for correcting my negated statement! Was wondering, if i replace my old negated statement with yours, would by proof by contradiction using intermediate value theorem work then? $\endgroup$ – Jayne Leblanc Sep 25 '16 at 11:30
  • $\begingroup$ You don't need the intermediate value theorem for this, see my comment to your question, and apply, for example. Bolzano-Weierstrass. $\endgroup$ – SamM Sep 25 '16 at 11:37
  • $\begingroup$ Hi @SamM! Thank you so much for your help! As my school's syllabus has not taught the above theorem you mentioned, I'm actually not allowed to use it. Hope you understand! $\endgroup$ – Jayne Leblanc Sep 25 '16 at 11:47
  • $\begingroup$ OK, have to learnt the extreme value theorem? If so, then look at the minimum of the function: if this is non-positive then you can arrive at a contradiction that way. $\endgroup$ – SamM Sep 25 '16 at 11:51
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Unfortunately, you negated incorrectly. The correct negation would be:

For all $c > 0$, there is some $x \in [a, b]$ such that $-c \leq f(x) \leq c$.


Let's prove the statement directly instead. Observe that for all $x \in [a, b]$, either $f(x) > 0$ or $f(x) < 0$. We can't have both positive parts and negative parts in the interval, since by the Intermediate Value Theorem this would imply that $f(d) = 0$ for some $d \in [a, b]$.

Without loss of generality, we may assume that $f(x) > 0$ for all $x \in [a, b]$. [For the other case, we could apply the same argument to $g(x) = -f(x)$.] Now since $f$ is continuous on $[a, b]$, it follows by the Extreme Value Theorem that there is some $z \in [a, b]$ such that $f(x) \geq f(z)$ for all $x \in [a, b]$. But since $f(z) > 0$, we may take: $$ c = \frac{f(z)}{2} $$ so that for all $x \in [a, b]$, we have that: $$ 0 < c < f(z) \leq f(x) $$

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